# 1.2 2d dft

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This module extends the ideas of the Discrete Fourier Transform (DFT) into two-dimensions, which is necessary for any image processing.

## 2d dft

To perform image restoration (and many other useful image processing algorithms) in a computer, we need a FourierTransform (FT) that is discrete and two-dimensional.

$F(k, l)=(u=\frac{2\pi k}{N}, v=\frac{2\pi l}{N}, F(u, v))$
for $k=\{0, , N-1\}$ and $l=\{0, , N-1\}$ .
$F(u, v)=\sum \sum f(m, n)e^{-(ium)}e^{-(ivm)}$
$F(k, l)=\sum_{m=0}^{N-1} \sum_{n=0}^{N-1} f(m, n)e^{-i\frac{2\pi km}{N}}e^{-i\frac{2\pi ln}{N}}$
where the above equation ( ) has finite support for an $N$ x $N$ image.

## Inverse 2d dft

As with our regular fourier transforms, the 2D DFT also has an inverse transform that allows us to reconstruct an imageas a weighted combination of complex sinusoidal basis functions.

$f(m, n)=\frac{1}{N^{2}}\sum_{k=0}^{N-1} \sum_{l=0}^{N-1} F(k, l)e^{\frac{i\times 2\pi km}{N}}e^{\frac{i\times 2\pi ln}{N}}$

## 2d dft and convolution

The regular 2D convolution equation is

$g(m, n)=\sum_{k=0}^{N-1} \sum_{l=0}^{N-1} f(k, l)h(m-k, n-l)$

Below we go through the steps of convolving two two-dimensional arrays. You can think of $f$ as representing an image and $h$ represents a PSF, where $h(m, n)=0$ for $m\land n> 1$ and $m\land n< 0$ . $h=\begin{pmatrix}h(0, 0) & h(0, 1)\\ h(1, 0) & h(1, 1)\\ \end{pmatrix}$ $f=\begin{pmatrix}f(0, 0) & & f(0, N-1)\\ & & \\ f(N-1, 0) & & f(N-1, N-1)\\ \end{pmatrix}$ Step 1 (Flip $h$ ):

$h(-m, -n)=\begin{pmatrix}h(1, 1) & h(1, 0) & 0\\ h(0, 1) & h(0, 0) & 0\\ 0 & 0 & 0\\ \end{pmatrix}$
Step 2 (Convolve):
$g(0, 0)=h(0, 0)f(0, 0)$
We use the standard 2D convolution equation ( ) to find the first element of our convolved image. In order to better understand what ishappening, we can think of this visually. The basic idea is to take $h(-m, -n)$ and place it "on top" of $f(k, l)$ , so that just the bottom-right element, $h(0, 0)$ of $h(-m, -n)$ overlaps with the top-left element, $f(0, 0)$ , of $f(k, l)$ . Then, to get the next element of our convolved image, we slide the flipped matrix, $h(-m, -n)$ , over one element to the right and get the following result: $g(0, 1)=h(0, 0)f(0, 1)+h(0, 1)f(0, 0)$ We continue in this fashion to find all of the elements ofour convolved image, $g(m, n)$ . Using the above method we define the general formula to find a particular element of $g(m, n)$ as:
$g(m, n)=h(0, 0)f(m, n)+h(0, 1)f(m, n-1)+h(1, 0)f(m-1, n)+h(1, 1)f(m-1, n-1)$

## Circular convolution

What does $H(k, l)F(k, l)$ produce?

2D Circular Convolution

$\stackrel{~}{g}(m, n)=\mathrm{IDFT}(H(k, l)F(k, l))=\mathrm{circularconvolutionin2D}$

Due to periodic extension by DFT ( ):

Based on the above solution, we will let

$\stackrel{~}{g}(m, n)=\mathrm{IDFT}(H(k, l)F(k, l))$
Using this equation, we can calculate the value for each position on our final image, $\stackrel{~}{g}(m, n)$ . For example, due to the periodic extension of the images, when circular convolution is applied we willobserve a wrap-around effect.
$\stackrel{~}{g}(0, 0)=h(0, 0)f(0, 0)+h(1, 0)f(N-1, 0)+h(0, 1)f(0, N-1)+h(1, 1)f(N-1, N-1)$
Where the last three terms in are a result of the wrap-around effect caused by the presence of the images copies located all around it.

## Zero padding

If the support of $h$ is $M$ x $M$ and $f$ is $N$ x $N$ , then we zero pad $f$ and $h$ to $M+N-1$ x $M+N-1$ (see ).

Circular Convolution = Regular Convolution

## Computing the 2d dft

$F(k, l)=\sum_{m=0}^{N-1} \sum_{n=0}^{N-1} f(m, n)e^{-i\frac{2\pi km}{N}}e^{-i\frac{2\pi ln}{N}}$
where in the above equation, $\sum_{n=0}^{N-1} f(m, n)e^{-i\frac{2\pi ln}{N}}$ is simply a 1D DFT over $n$ . This means that we will take the 1D FFT of each row; if wehave $N$ rows, then it will require $N\lg N$ operations per row. We can rewrite this as
$F(k, l)=\sum_{m=0}^{N-1} {f}^{}(m, l)e^{-i\frac{2\pi km}{N}}$
where now we take the 1D FFT of each column, which means that if we have $N$ columns, then it requires $N\lg N$ operations per column.
Therefore the overall complexity of a 2D FFT is $O(N^{2}\lg N)$ where $N^{2}$ equals the number of pixels in the image.

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Source:  OpenStax, Intro to digital signal processing. OpenStax CNX. Jan 22, 2004 Download for free at http://cnx.org/content/col10203/1.4
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